Find the area of the region that lies inside the graph of the first polar equation and outside of the graph of the second polar equation. r = 1 + cos(𝜃), r = 2 − cos(𝜃)

Answer 1

Intersections: 1 + cosθ = 2 - cosθ; cosθ = 1/2; θ = π/3 or θ = - π/3;

A = ∫₀^π/3[(1 + cosθ)² - (2 - cosθ)²]dθ = ∫₀^π/3(6cosθ - 3)dθ = (6sinθ - 3θ)₀^π/3 = 6sinπ/3 - π = 3√3 - π